Using y, x, and a, write a general equation for the parabola whose vertex is at the origin and whose axis of symmetry is the x-axis.
My answers were y^2=x or y^2=-x.
What's the real answer? Or did I get it right.
2006-11-21 13:54:43 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
both are right, they give two different parabolas that are both answers.
2006-11-21 13:59:06 · answer #1 · answered by Anonymous · 0⤊ 1⤋
your answers do not include a.
try, y = ax^2, a is a real number and is not 0 (if a=0 then we do not have a parabola).
if the vertex is at the origin, then (0,0) must be on the parabola.
if y = (x+a)^2, then if x=0, y = a^2, not necessarily 0 (only if a = 0).
hope this helps...
2006-11-21 17:38:10 · answer #2 · answered by mr green 4 · 0⤊ 0⤋
Looks good to me. There are two parabolas that match the description, opening in opposite directions. Stylistically, you should put that negative sign with the y^2. It's the same thing but it's traditional.
2006-11-21 13:59:34 · answer #3 · answered by modulo_function 7 · 0⤊ 1⤋
y=(x+a)^2
2006-11-21 13:59:33 · answer #4 · answered by yupchagee 7 · 0⤊ 1⤋
the mother function is y=x^2 and the inverse of that would make it symmetry to the x-axis
so find the inverse of y=x^2
and ur answer is right!
2006-11-21 14:01:46 · answer #5 · answered by Kevin C 1 · 0⤊ 1⤋
EITHER IS GOOD direction doesn't matter
2006-11-21 14:02:32 · answer #6 · answered by Brian D 5 · 0⤊ 0⤋
YOU ARE RIGHT !!!
GOOD JOB !!!
2006-11-21 14:00:08 · answer #7 · answered by ilovemath_pi 2 · 0⤊ 1⤋